题目:
本题还有另外一种思路,可以参考leetcode详解.pdf
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
解题思路:该题与Search in Rotated Sorted Array的主要区别在于需要考虑A[low]与A[mid]相等、A[mid]与A[high]相等以及A[low]A[mid]A[high]三者相等的情形。若A[low]与A[mid]相等而A[mid]与A[high]不等,则搜索右边序列;若A[mid]与A[high]相等而A[low]与A[mid]不等,则搜索左边序列;若A[low]A[mid]A[high]三者相等,则在最坏情况下需要对两边的序列进行搜索。
给出递归程序:
class Solution { public: bool search(int A[], int n, int target) { int high,low; low=0; high=n-1; return Bsearch(A,low,high,target); } private: bool Bsearch(int A[], int low, int high, int target){ if((high-low)==0){ return A[low]==target; }else if((high-low)==1){ return (A[low]==target)||(A[high]==target); } int mid=(low+high)/2; if(A[mid]==target)return true; if(A[low]<A[mid]){ if((A[low]<=target)&&(A[mid]>target)){ return Bsearch(A,low,mid,target); }else if(A[mid]==A[high]){ return false; }else{ return Bsearch(A,mid,high,target); } }else if(A[mid]<A[high]){ if((A[mid]<target)&&(A[high]>=target)){ return Bsearch(A,mid,high,target); }else if(A[low]==A[mid]){ return false; }else{ return Bsearch(A,low,mid,target); } }else if(A[low]!=A[mid]){ return Bsearch(A,low,mid,target); }else if(A[mid]!=A[high]){ return Bsearch(A,mid,high,target); }else{ return Bsearch(A,low,mid,target)||Bsearch(A,mid,high,target); } } };
本题还有另外一种思路,可以参考leetcode详解.pdf